Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO 2 (s) + H 2 SO 4 (aq) ⟶ BaSO 4 (s) + H 2 O 2 (aq) BaO2 (s) + H2SO4 (aq) ⟶BaSO4 (s) + H2O2 (aq) How many milliliters of 3.00 M H 2 SO 4 (aq) 3.00 M H2SO4 (aq) are needed to react completely with 92.5 g BaO 2 (s) ? 92.5 g BaO2 (s) ?

There is 182 mL of 3.00 M H2SO4 needed

Explanation:

Step 1: Data given

Mass of BaO2 = 92.5 grams

Molarity of 3.00 M H2SO4

Step 2: The balanced equation

BaO2 (s) + H2SO4 (aq) ⟶ BaSO4 (s) + H2O2 (aq)

For 1 mole of BaO2 we need 1 mole of H2SO4 to produce 1 mole of BaSO4 and 1 mole H2SO2

Step 3: Calculate moles of BaO2

Moles BaO2 = mass BaO2 / Molar mass BaO2

Moles BaO2 = 92.5 / 169.33 g/mol

Moles BaO2 = 0.546 moles

Step 4: Calculate moles of H2SO4

For 1 mole BaO2 we need 1 mole of H2SO4

For 0.546 moles of BaO2 we need 0.546 moles of H2SO4

Step 5: Calculate the volume of H2SO4 needed

Volume = moles / molarity

Volume = 0.546 mol / 3.00 mol/L

Volume = 0.182 L = 182 mL

There is 182 mL of 3.00 M H2SO4 needed